Question

Let O be the vertex and Q be any point on the parabola ${{x}^{2}}=8y$. If the point P divides the line segment OQ internally in the ratio $1:3$, then the locus of P is

& A){{x}^{2}}=y \\\ & B){{y}^{2}}=x \\\ & C){{y}^{2}}=2x \\\ & D){{x}^{2}}=2y \\\ \end{aligned}$$

Answer 1

We know that the vertex of the parabola ${{x}^{2}}=4ay$is $O\left( 0,0 \right)$. We know that the general point on the parabola ${{x}^{2}}=4ay$ is $Q\left( 2at,a{{t}^{2}} \right)$. We know that if $P\left( {{x}_{1}},{{y}_{1}} \right)$ and $Q\left( {{x}_{2}},{{y}_{2}} \right)$ are two points, then $R\left( {{x}_{3}},{{y}_{3}} \right)$ is divided by $P\left( {{x}_{1}},{{y}_{1}} \right)$ and $Q\left( {{x}_{2}},{{y}_{2}} \right)$ in the ratio $m:n$internally, then $R\left( {{x}_{3}},{{y}_{3}} \right)$ is equal to $R\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right)$. By using this concept, we can find the equation of parabola.

Complete step-by-step answer:
We know that the vertex of the parabola ${{x}^{2}}=4ay$is $O\left( 0,0 \right)$.From the question, we were given that O is the vertex of the parabola.
Now we should compare ${{x}^{2}}=8y$ with ${{x}^{2}}=4ay$.

& \Rightarrow 4a=8 \\\ & \Rightarrow a=2....(1) \\\ \end{aligned}$$ So, from equation (1) we can write the vertex of the parabola $${{x}^{2}}=8y$$ is $$O\left( 0,0 \right)$$. We know that the general point on the parabola $${{x}^{2}}=4ay$$ is $$Q\left( 2at,a{{t}^{2}} \right)$$. So, from equation (1), we can write that the general point on the parabola $${{x}^{2}}=8y$$ is $$Q\left( 4t,2{{t}^{2}} \right)$$. We know that if $$P\left( {{x}_{1}},{{y}_{1}} \right)$$ and $$Q\left( {{x}_{2}},{{y}_{2}} \right)$$ are two points, then $$R\left( {{x}_{3}},{{y}_{3}} \right)$$ is divided by $$P\left( {{x}_{1}},{{y}_{1}} \right)$$ and $$Q\left( {{x}_{2}},{{y}_{2}} \right)$$ in the ratio $$m:n$$internally, then $$R\left( {{x}_{3}},{{y}_{3}} \right)$$ is equal to $$R\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right)$$.  From the question, we were given that the point $$P(x,y)$$ divides the line segment OQ internally in the ratio $$1:3$$. $$\begin{aligned} & \Rightarrow P(x,y)=P\left( \dfrac{\left( 1 \right)(4t)+(3)(0)}{1+3},\dfrac{(1)(2{{t}^{2}})+(3)(0)}{1+3} \right) \\\ & \Rightarrow P(x,y)=P\left( \dfrac{4t}{4},\dfrac{2{{t}^{2}}}{4} \right) \\\ & \Rightarrow P(x,y)=P\left( t,\dfrac{{{t}^{2}}}{2} \right) \\\ \end{aligned}$$ We know that the x-coordinate of $$P(x,y)$$ is equal to x and the y-coordinate of $$P(x,y)$$ is equal to y. $$\begin{aligned} & x=t....(2) \\\ & y=\dfrac{{{t}^{2}}}{2}.....(3) \\\ \end{aligned}$$ Now let us substitute equation (2) in equation (3), then $$\begin{aligned} & y=\dfrac{{{x}^{2}}}{2} \\\ & \Rightarrow {{x}^{2}}=2y......(4) \\\ \end{aligned}$$ **So, the correct answer is “Option D”.** **Note:** Students may have a misconception that if $$P\left( {{x}_{1}},{{y}_{1}} \right)$$ and $$Q\left( {{x}_{2}},{{y}_{2}} \right)$$ are two points, then $$R\left( {{x}_{3}},{{y}_{3}} \right)$$ is divided by $$P\left( {{x}_{1}},{{y}_{1}} \right)$$ and $$Q\left( {{x}_{2}},{{y}_{2}} \right)$$ in the ratio $$m:n$$ internally, then $$R\left( {{x}_{3}},{{y}_{3}} \right)$$ is equal to $$R\left( \dfrac{m{{x}_{2}}-n{{x}_{1}}}{m-n},\dfrac{m{{y}_{2}}-n{{y}_{1}}}{m-n} \right)$$. But we know that if $$P\left( {{x}_{1}},{{y}_{1}} \right)$$ and $$Q\left( {{x}_{2}},{{y}_{2}} \right)$$ are two points, then $$R\left( {{x}_{3}},{{y}_{3}} \right)$$ is divided by $$P\left( {{x}_{1}},{{y}_{1}} \right)$$ and $$Q\left( {{x}_{2}},{{y}_{2}} \right)$$ in the ratio $$m:n$$, then $$R\left( {{x}_{3}},{{y}_{3}} \right)$$ is equal to $$R\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right)$$. So, this misconception should be avoided.